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thnx +tan3x=2tan2x find general solution. Please give me ans. Please give me ans thnx +tan3x=2tan2x find general solution. Please give me ans.Please give me ans
tan x + tan 2x = tan 3xNow let us simplify, tan x + tan 2x = tan 3xtan x + tan 2x – tan 3x = 0tan x + tan 2x – tan (x + 2x) = 0On using the formula, tan (A+B) = [tan A + tan B] / [1 – tan A tan B]Therefore, tan x + tan 2x – [[tan x + tan 2x]/[1- tan x tan 2x]] = 0(tan x + tan 2x) (1 – 1/(1- tan x tan 2x)) = 0(tan x + tan 2x) ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0Now, (tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0(tan x + tan 2x) = 0 or [– tan x tan 2x] = 0 tan x = tan (-2x) or -tan x tan 2x = 0tan x = tan (-2x) or 2tan2 x / (1 – tan2 x) = 0[Using, tan 2x = 2 tan x / 1-tan2 x] x = nπ + (-2x)orx = mπ + 0 3x = nπorx = mπ x = nπ/3 or x = mπ∴ the general solution is x = nπ/3 or mπ,
tanx + tan 3x = 2tan 2x tanx + tan3x = tan2x +tan2xthnx - tan2x = tan2x - tan3xtam 2x -tanx =tan 3x - tan2xtan( 2x-x )( 1 + tan2x tanx ) = tan(3x-2x) ( 1 + tan 3x tan2x)tanx ( 1+ tan2x tanx ) - tanx ( 1+tan3xtan2x)=0tanx {1 + tan2xtanx -1-tan3xtan2x}=0tanx (tan2xtanx- tan3xtan2x)=0tanx tan2x( tanx -tan3x) =0If tanx = 0 or tan2x = 0 x = nπ. tan2x = nπ x=nπ/2and tanx - tan3x =0 tanx = tan3x tan3x = tanx 3x = nπ +x 2x= nπx = nπ /2
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